We first observe that Lemma REF  applies to \(T_{g_1^\pm },T_{g_2},T_{g_3}\)  and \(T_{g_4}\)  since the multiplication by \(\mathop {\mathrm {sgn}}\nolimits x\)  is bounded on \(L^p(w_p)\)  for all \(1\le p<\infty \)  and on \(L^{1,\infty }(w_1)\) .
To prove the lemma, it is enough to show \(T_{g_1^\pm },T_{g_2},T_{g_3},T_{g_4}\in \mathbb {B}(\mathcal {H}^1({\mathbb {R}}),L^1({\mathbb {R}}))\) . Indeed, if we write \(g_1^\pm =g_{1,a,b}^\pm \)  and \(g_{j}=g_{j,a,b}\) , then it is easy to see that
\(\overline{g_{1,a,b}^\pm (y,x)}=\overline{a} k_1^+(x,y)\mp \overline{a}k_1^-(x,y)+i\overline{b} k_2^+(x,y)\mp i\overline{b}k_2^-(x,y)=g^\mp _{1,\overline{a},i\overline{b}}(x,y).\) 

Hence \((T_{g_1^\pm ,a,b})^*=T_{g^\mp _{1,\overline{a},i\overline{b}}}\) . Similarly, we have \((T_{g_{2,a,b}})^*=T_{g_{3,\overline{a},i\overline{b}}}\)  and \((T_{g_{3,a,b}})^*=T_{g_{2,\overline{a},i\overline{b}}}\) . Therefore, since \(({\mathcal {H}}^1)^*={\mathop {\mathrm {BMO}}}\)  (see [1]}), if \(T_{g_1^\pm },T_{g_2},T_{g_3}\in \mathbb {B}(\mathcal {H}^1({\mathbb {R}}),L^1({\mathbb {R}}))\)  for any \(a,b\in {\mathbb {C}}\) , then \(T_{g_1^\pm },T_{g_2},T_{g_3}\in \mathbb {B}(L^\infty ({\mathbb {R}}),{\mathop {\mathrm {BMO}}}({\mathbb {R}}))\)  by the duality argument.
The proof of \(T_{g_1^\pm },T_{g_2},T_{g_3},T_{g_4}\in \mathbb {B}(\mathcal {H}^1({\mathbb {R}}),L^1({\mathbb {R}}))\)  follows a classical argument in the proof of the \(\mathcal {H}^1\) -\(L^1\)  boundedness of Calderón–Zygmund operators.
We let \(f\in \mathcal {H}^1\)  and apply the atomic decomposition (see [2]}) to obtain
\(f=\sum _{j=1}^\infty \lambda _ja_j(x),\quad \sum _{j=1}^\infty |\lambda _j|\lesssim {\Vert f\Vert }_{\mathcal {H}^1},\) 
where \(\lambda _j\in {\mathbb {C}}\)  and \(a_j\)  are \(L^\infty \) -atoms for \(\mathcal {H}^1\)  satisfying, with some \(x_j\in {\mathbb {R}}\)  and \(r_j\ge 2\) ,
\(\mathop {\mathrm {supp}}\nolimits a_j\subset (x_j-r_j,x_j+r_j),\quad {\Vert a_j\Vert }_{L^\infty }\lesssim r_j^{-1},\quad \int a_j(x)dx=0.\) 
Hence, for a given integral operator \(T\) , once we obtain
\({\Vert T a_j\Vert }_{L^1({\mathbb {R}})}\lesssim 1\) 

uniformly in \(j\) , \(T\)  is bounded from \({\mathcal {H}}^1\)  to \(L^1\)  since
\({\Vert Tf\Vert }_{L^1({\mathbb {R}})}\le \sum _{j=1}^\infty |\lambda _j|{\Vert Ta_j\Vert }_{L^1({\mathbb {R}})}\lesssim {\Vert f\Vert }_{{\mathcal {H}}^1({\mathbb {R}})}.\) 
It is thus enough to prove [lemma24proof1]lemma24proof1 for \(T= T_{g_1^\pm },T_{g_2},T_{g_3}\)  and \(T_{g_4}\) .
Let \(I=(x_0-r,x_0+r)\)  with some fixed \(x_0\in {\mathbb {R}},r\ge 2\)  and take an \(L^\infty \) -atom \(a\)  satisfying
\(\mathop {\mathrm {supp}}\nolimits a\subset I,\quad {\Vert a\Vert }_{L^\infty }\lesssim r^{-1},\quad \int _Ia(x)dx=0.\) 
We let \(I_*=(x_0-3r,x_0+3r)\cup (-x_0-3r,-x_0+3r)\)  and decompose
\({\Vert Ta\Vert }_{L^1({\mathbb {R}})}={\Vert Ta\Vert }_{L^1(I_*)}+{\Vert Ta\Vert }_{L^1(I_*^c)}.\) 
For the first term, Lemma REF  and Hölder's inequality imply
\({\Vert Ta\Vert }_{L^1(I_*)}\le |I_*|^{1/2}{\Vert Ta\Vert }_{L^2(I_*)}\lesssim r^{1/2}{\Vert a\Vert }_{L^2(I_*)}\lesssim r^{1/2}{\Vert a\Vert }_{L^\infty }r^{1/2}\lesssim 1\) 

uniformly in \(x_0\)  and \(r\)  for all \(T= T_{g_1^\pm },T_{g_2},T_{g_3},T_{g_4}\) .
To deal with the second term, we let \(x\in I_*^c\)  and \(y\in \mathop {\mathrm {supp}}\nolimits a\subset I\) . Then
\(||x|\pm |y||\ge \min _\pm |x\pm y|\ge r\ge 2\) 

and hence \(\psi (\big ||x|\pm |y|\big |^2)=1\)  since \(\psi \equiv 1\)  on \([2,\infty )\) . It also follows that
\(|x\pm y|\ge |x\pm x_0|-|x_0-y|\ge |x\pm x_0|-r\ge |x\pm x_0|/2.\) 

With these remarks at hand, we compute
\(|g_1^\pm (x,y)-g_1^\pm (x,x_0)|&\lesssim \sum _\pm \left(\left|\frac{1}{|x|\pm |y|}-\frac{1}{|x|\pm |x_0|}\right|+\left|\frac{1}{|x|\pm i|y|}-\frac{1}{|x|\pm i|x_0|}\right|\right)\\&\lesssim \sum _\pm \left(\left|\frac{|x_0|-|y|}{(|x|\pm |y|)(|x|\pm |x_0|)}\right|+\left|\frac{|x_0|-|y|}{(|x|\pm i|y|)(|x|\pm i|x_0|)}\right|\right)\\&\lesssim \frac{|x_0-y|}{\displaystyle \min _\pm |x\pm x_0|^2}.\) 

Using the relations
\(\frac{\mathop {\mathrm {sgn}}\nolimits y}{|x|+|y|} -\frac{\mathop {\mathrm {sgn}}\nolimits y}{|x|-|y|}&=\frac{-2y}{x^2-y^2}=\frac{1}{x+y}-\frac{1}{x-y},\\\frac{\mathop {\mathrm {sgn}}\nolimits y}{|x|+i|y|}-\frac{\mathop {\mathrm {sgn}}\nolimits y}{|x|-i|y|}&=\frac{-2iy}{x^2+y^2}=\frac{1}{x+iy}-\frac{1}{x-iy},\) 

we also have, for \(x\in I_*^c\)  and \(y\in I\) ,
\(|g_3(x,y)-g_3(x,x_0)|&\lesssim \sum _\pm \left(\left|\frac{1}{x\pm y}-\frac{1}{x\pm x_0}\right|+\left|\frac{1}{x\pm iy}-\frac{1}{x\pm ix_0}\right|\right)\\&\lesssim \frac{|x_0-y|}{|x+x_0|^2}+\frac{|x_0-y|}{|x-x_0|^2}\lesssim \frac{|x_0-y|}{\displaystyle \min _\pm |x\pm x_0|^2}.\) 

Moreover, we also obtain, for \(x\in I_*^c\)  and \(y\in I\) ,
\(|g_2(x,y)-g_2(x,x_0)|+|g_4(x,y)-g_4(x,x_0)|\lesssim \frac{|x_0-y|}{\displaystyle \min _\pm |x\pm x_0|^2}.\) 

Hence, for all the cases \(K=g_1^\pm ,g_2,g_3,g_4\) , \(T_K\)  satisfies
\(\nonumber {\Vert T_Ka\Vert }_{L^1(I_*^c)}&=\int _{I_*^c}\left|\int _I\Big (K(x,y)-K(x,x_0)\Big )a(y)dy\right|dx\\\nonumber &\lesssim {\Vert a\Vert }_{L^\infty } \int _I |x_0-y|dy \int _{I_*^c}\frac{1}{\displaystyle \min _\pm |x\pm x_0|^2} dx\\&\lesssim r\int _{I_*^c}\frac{1}{\displaystyle \min _\pm |x\pm x_0|^2} dx.\) 

Setting \(U_{x_0}=\lbrace x\ |\ |x+x_0|\le |x-x_0|\rbrace \) , we further obtain
\(\int _{I_*^c}\frac{1}{\displaystyle \min _\pm |x\pm x_0|^2} dx&=\int _{I_*^c\cap U_{x_0}}\frac{1}{|x+ x_0|^2} dx+\int _{I_*^c\cap U_{x_0}^c}\frac{1}{|x-x_0|^2} dx\\&\le \int _{|x+x_0|\ge 3r}\frac{1}{|x+ x_0|^2} dx+\int _{|x-x_0|\ge 3r}\frac{1}{|x-x_0|^2} dx\lesssim r^{-1}.\) 

It thus follows that
\({\Vert T_Ka\Vert }_{L^1(I_*^c)}\lesssim 1\)  uniformly in \(x_0,r\) . This bound, together with [lemma24proof2]lemma24proof2, implies [lemma24proof1]lemma24proof1. This completes the proof.
